A projectile is thrown horizontally at 10.0 m/s.The projectile hits the ground 0.510 s later.What is the angle of impact the projectile makes with the horizontal ground?

Question

Quizplus · Accepted Answer

Since the projectile is thrown horizontally, its initial vertical velocity is zero. We can use the kinematic equation:

y = 1/2 * a * t^2

where y is the vertical displacement, a is the acceleration due to gravity (-9.8 m/s^2), and t is the time. Since the projectile hits the ground after 0.510 s, we can substitute these values into the equation:

0 = 1/2 * (-9.8) * (0.510)^2

Solving for y, we get:

y = 1.25 m

The horizontal displacement can be found using the kinematic equation:

x = v * t

where x is the horizontal displacement and v is the initial horizontal velocity (10.0 m/s). Substituting the values:

x = 10.0 * 0.510

x = 5.10 m

Now, we can find the angle of impact using the trigonometric equation:

tan(theta) = y / x

Substituting the values we found:

tan(theta) = 1.25 / 5.10

theta = tan^-1(0.245)

theta = -26.6° (Note: the angle is negative because it is measured below the horizontal)

A Projectile Is Thrown Horizontally at 10 °\degree° B)-266 °\degree° C)-270 °\degree° D)-600 °\degree°

A Projectile Is Thrown Horizontally at 10 $\degree$ B)-266 $\degree$ C)-270 $\degree$ D)-600 $\degree$